Answer
a) $(d P)_v=1.339kPa$
b) $(d P)_T=-1.339kPa$
c) $dP=0$
Work Step by Step
An ideal gas equation can be expressed as $P=R T / v$. Noting that $R$ is a constant and $P=P(T, v)$,
$$
d P=\left(\frac{\partial P}{\partial T}\right)_v d T+\left(\frac{\partial P}{\partial v}\right)_T d v=\frac{R d T}{v}-\frac{R T d v}{v^2}
$$ (a) The change in $T$ can be expressed as $d T \cong \Delta T=350 \times 0.01=3.5 \mathrm{~K}$. At $v=$ constant,
$$
(d P)_v=\frac{R d T}{v}=\frac{\left(0.287 \mathrm{kPa} \cdot \mathrm{m}^3 / \mathrm{kg} \cdot \mathrm{K}\right)(3.5 \mathrm{~K})}{0.75 \mathrm{~m}^3 / \mathrm{kg}}=1.339 \mathrm{kPa}
$$ (b) The change in $v$ can be expressed as $d v \cong \Delta v=0.75 \times 0.01=0.0075 \mathrm{~m}^3 / \mathrm{kg}$. At $T=$ constant, $$
(d P)_T=-\frac{R T d v}{v^2}=-\frac{\left(0.287 \mathrm{kPa} \cdot \mathrm{m}^3 / \mathrm{kg} \cdot \mathrm{K}\right)(300 \mathrm{~K})\left(0.0075 \mathrm{~m}^3 / \mathrm{kg}\right)}{\left(0.75 \mathrm{~m}^3 / \mathrm{kg}\right)^2}=-\mathbf{1 . 3 3 9 k P a}
$$ (c) When both $v$ and $T$ increases by $1 \%$, the change in $P$ becomes
$$ d P=(d P)_v+(d P)_T=1.339+(-1.339)=\mathbf{0}
$$ Thus the changes in $T$ and $v$ balance each other.
